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3 years ago

A 75-kilogram hockey player is skating across the

1 answer:

azamat3 years ago

6 0

I think it is (3) 690 N. You can use the theorem of Impulse and momentum where the impulse is equal to the change in momentum or:
$F \Delta t=m v_{f}-m v_{i}$ (where f is for final and i for initial).
So you get:
$F*0.65=0-(75*6)$
$F=692 N$

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A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.

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In this problem we have the electric field intensity E:

E = 6.5 × $10^4$ newtons/coulomb

We have the magnitude of the load:

q = 6.4 × $10 ^{-19}$ coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × $10^{-2}$ meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × $10^{-19}$ )(6.5 × $10^4$ )(1.2 × $10^{-2}$ )

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3 years ago

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3 years ago

Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years i

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Answer:

r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

F = ma

G m M / r² = m a

The centripetal acceleration is given by

a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

v = d / t

The distance traveled Esla orbits, in a circle the distance is

d = 2 π r

Time in time to complete the orbit, called period

v = 2π r / T

Let's replace

G m M / r² = m a

G M / r² = (2π r / T)² / r

G M / r² = 4π² r / T²

G M T² = 4π² r3

r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

T = 1.03 10⁸ s

Let's calculate

r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

r = ∛ (21.44 10³⁵ / 39.478)

r = ∛(0.0543087 10 36)

r = 0.3787 10¹² m

r = 3.787 10¹¹ m

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3 years ago